A free response question involving inductance was posted on 23^{rd} December 2007 for your practice. This was the question:

_{1}, R

_{2}and R

_{3}are i

_{1}, i

_{2}and i

_{3}respectively. The switch S is closed at t = 0.

(a) Calculate the currents i_{1}, i_{2} and i_{3} immediately after closing the switch S (when t = 0).

(b) Calculate the currents i_{1}, i_{2} and i_{3} when steady state is reached.

(c) Show the nature of variation of the current through the battery graphically, with time ‘t’ on the X-axis and the current (i_{1}) on the Y-axis.

(d) What are the values of the slopes of the above graph at t = 0 and at t = ∞.

(e) The switch S is opened and the resistance R_{3} is disconnected. If the switch S is now closed, what time is required for the current to attain 63.2% of the final steady value?

As promised, I give below the answer:

**(a)** Since the parallel branch containing R_{3} does not contain any inductance, the current through that branch (i_{3}) rises to the maximum possible value in this condition immediately on closing the switch S. The same current now flows through R_{1} since the initial current i_{2} ( at t = 0) through the branch containing the inductance L and the resistance R_{2} is zero. [Note that in an LR circuit, the current I at the instant ‘t’ is given by **I = I _{0}(1**–

**e**where I

^{–Rt/L})_{0}is the

*fina*l steady current]

Therefore, when t = 0, i_{1 }= i_{3} = (12 V)/ (8 Ω + 8 Ω ) = 0.75 A and i_{2} = 0

**(b)** Steady state is reached when t = ∞. In this condition, the current is controlled by the resistances only [since dI/dt is zero and no opposing induced voltage L(dI/dt) is developed across the inductance]. The circuit then behaves as made of the parallel combination of R_{2 }and_{ }R_{3} (having effective value of 4 Ω) in series with the battery and the resistance R_{1}. The effective resistance of the circuit is (8 + 4) Ω = 12 Ω and the current delivered by the battery (i_{1}) is (12 V)/ (12 Ω) = 1 A.

_{2}= i

_{3}= 0.5 A.

**(c) **The variation of the current i_{1 }with time t is shown in the adjoining graph. [The current grows exponentially in the inductive branch and this is why the current delivered by the battery finally rises to the steady value of 1 A as shown].

**(d) **The slope of the graph when t = o is infinity since the current grows abruptly to 0.75 A because of the purely resistive parallel branch containing R_{3}. The slope when t = ∞ is zero since the current has settled to the final steady value.

**(e) **When R_{3 }is disconnected, We have a series LR circuit with L = 2 H and R = R_{1} + R_{2} = 16 Ω. The time constant of the circuit is L/R = 2/16 = **0.125 s**. The current grows to 63.2 % of the final steady current within this time.

[Time constant is L/R which is the time required for the current to become (1 – 1/e) times the final steady current. This is obtained by putting t = L/R in the equation,

I = I_{0}(1– e^{–Rt/L})].

Merry Christmas!

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